Silicate, among the following, has only one oxygen atom of \([SiO_4]^{4-}\) shared with the next unit:

1.	Sheet silicate	2.	Pyrosilicate
3.	Three dimensional silicate	4.	Linear chain silicate

The incorrect statement among the following is:

1.  SnF₄ is ionic in nature.

2.  PbF₄ is covalent in nature.

3.  SiCl₄ is easily hydrolysed.

4.  GeX₄ (X = F, Cl, Br, I) is more stable than GeX₂

Hydrated aluminium chloride is an ionic compound and soluble in water giving -

1. Al³⁺ and Cl^- ions

2. ${\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}$ and Cl^- ions

3. ${\left[{\mathrm{AlCl}}_2{\left({\mathrm{H}}_2\mathrm{O}\right)}_4\right]}^{+}$ and ${\left[{\mathrm{AlCl}}_4{\left({\mathrm{H}}_2\mathrm{O}\right)}_2\right]}^{-}$  ions

4. None of the above

${\mathrm{Al}}_4{\mathrm{C}}_3$ is an ionic carbide. It is also known as :

1. Acetylide

2. Methanide

3. Allyide

4. Alloy

Anhydrous aluminium chloride fumes in moist air owing to the formation of :

1. gaseous aluminium chloride

2. chlorine

3. chlorine dioxide

4. hydrogen chloride

The silicate anion in the mineral kinolte is a chain of three SiO₄ tetrahedra, that share corners with adjacent tetrahedra. The charge of the silicate anion is :

1. -4

2. -8

3. -6

4. -2

A compound that is sparingly soluble in cold water and fairly soluble in hot water is-

1. $\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$

2. ${\mathrm{PbCl}}_2$

3. ${\mathrm{PbSO}}_4$

4. ${\mathrm{PbCrO}}_4$

Concentrated nitric acid reacts with iodine to give :

1. HOI

2. HI

3. HOIO₂

4. HOIO₃

The dipole moment of ${\mathrm{H}}_2{\mathrm{O}}_2$ is more than that of ${\mathrm{H}}_2\mathrm{O}$ but ${\mathrm{H}}_2{\mathrm{O}}_2$ is not a good solvent because :

1. It has a very high dielectric constant so that ionic compounds cannot be dissolved in it.

2. It does not act as an oxidising agent

3. It acts as a reducing agent

4. It dissociates easily and acts as an oxidising agent in chemical reactions 

The dissolution of $\mathrm{Al}{\left(\mathrm{OH}\right)}_3$ by a solution of NaOH results in the formation of :

1. ${\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4{\left(\mathrm{OH}\right)}_2\right]}^{+}$

2. $\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_3{\left(\mathrm{OH}\right)}_3\right]$

3. ${\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_2{\left(\mathrm{OH}\right)}_4\right]}^{-}$

4. $\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{\left(\mathrm{OH}\right)}_3\right]$